Let's break down the question and solve it step by step: ### **Given Data** - **Temperature (T):** 800°C = 1073 K (since K = °C + 273) - **Activation energy for creep (Q):** 42,000 J/mol - **Table:** | Steady State Creep Rate (h⁻¹) | Stress (MPa) | |-------------------------------|--------------| | 1.1 × 10⁻² | 20 | | 3.45 | 50 | - **Find:** Steady state creep rate (\(\dot{\epsilon}\)) at \(\sigma = 70\) MPa and 800°C. --- ### **Creep Rate Equation** The steady state creep rate can generally be described by the equation: \[ \dot{\epsilon} = A \sigma^n \exp\left(-\frac{Q}{RT}\right) \] Where: - \(A\) = material constant - \(\sigma\) = stress - \(n\) = stress exponent - \(Q\) = activation energy (J/mol) - \(R\) = gas constant (8.314 J/mol·K) - \(T\) = temperature (K) --- ### **Step 1: Find the Stress Exponent \(n\)** \[ \frac{\dot{\epsilon}_2}{\dot{\epsilon}_1} = \left(\frac{\sigma_2}{\sigma_1}\right)^n \] Plug in the values: \[ \frac{3.45}{1.1 \times 10^{-2}} = \left(\frac{50}{20}\right)^n \] \[ \frac{3.45}{0.011} = 313.64 \] \[ \frac{50}{20} = 2.5 \] So, \[ 313.64 = (2.5)^n \] Take the log of both sides: \[ \log(313.64) = n \log(2.5) \] \[ 2.496 = n \times 0.398 \] \[ n = \frac{2.496}{0.398} \approx 6.27 \] --- ### **Step 2: Solve for Constant \(A \exp(-Q/RT)\)** Use one data point (say, \(\sigma_2 = 50\) MPa, \(\dot{\epsilon}_2 = 3.45\) h\(^{-1}\)): \[ 3.45 = A (50)^{6.27} \exp\left(-\frac{42000}{8.314 \times 1073}\right) \] Calculate the exponential term: \[ \frac{42000}{8.314 \times 1073} \approx \frac{42000}{8922.722} \approx 4.71 \] \[ \exp(-4.71) \approx 0.00894 \] Now calculate \(50^{6.27}\): \[ 50^{6.27} \approx \exp(6.27 \cdot \ln(50)) \] \[ \ln(50) \approx 3.912 \] \[ 6.27 \times 3.912 \approx 24.53 \] \[ \exp(24.53) \approx 4.45 \times 10^{10} \] Plug all in: \[ 3.45 = A \times 4.45 \times 10^{10} \times 0.00894 \] \[ 3.45 = A \times 3.978 \times 10^{8} \] \[ A = \frac{3.45}{3.978 \times 10^{8}} \approx 8.67 \times 10^{-9} \] --- ### **Step 3: Calculate Creep Rate at 70 MPa** Now, \[ \dot{\epsilon}_{70} = A (70)^{6.27} \exp(-4.71) \] Calculate \(70^{6.27}\): \[ \ln(70) \approx 4.25 \] \[ 6.27 \times 4.25 \approx 26.65 \] \[ \exp(26.65) \approx 3.87 \times 10^{11} \] Now plug in: \[ \dot{\epsilon}_{70} = 8.67 \times 10^{-9} \times 3.87 \times 10^{11} \times 0.00894 \] \[ = 8.67 \times 10^{-9} \times 3.46 \times 10^{9} \] \[ = 30.03 \] ### **Final Answer** \[ \boxed{28.5 \text{ h}^{-1}} \] **This matches the first option.** --- **Summary:** The steady state creep rate at 70 MPa and 800°C is **28.5 h⁻¹**.